Respuesta :
Answer:
Option (4).
Step-by-step explanation:
Given question is incomplete; here is the complete question.
Jefferson is plotting the vertices of an isosceles right triangle on a coordinate graph. He has plotted the two points shown. Where should he plot the third point?
The point is on the graph is (-2,-4) (4,-4)
The options are
(-2,-9)
(3,1)
(-2,0)
(4,2)
Distance between (-2, -4) and (4, -4) = 6 units
Now we will check every option,
1). For point (-2, -9),
Distance between (-2, -4) and (-2, -9) = [tex]\sqrt{(-2+2)^2+(-4+9)^2}=5[/tex]
Similarly, distance between (4, -4) and (-2, -9) = [tex]\sqrt{(4+2)^2+(-4+9)^2}=\sqrt{61}[/tex]
None side is equal. So it's not an isosceles triangle.
2). For a point (3, 1),
Distance between (3, 1) and (-2, -4) = [tex]\sqrt{(3+2)^2+(1+4)^2}=5\sqrt{2}[/tex]
Distance between (3, 1) and (4, -4) = [tex]\sqrt{(3-4)^2+(1+4)^2}=\sqrt{26}[/tex]
Since none side of the triangle are not equal, triangle is not an isosceles triangle.
3). For (-2,0),
Distance between (-2, 0) and (-2, -4) = 4
Distance between (-2, 0) and (4, -4) = [tex]\sqrt{(-2-4)^2+(0+4)^2}[/tex] = [tex]2\sqrt{13}[/tex]
None side is equal so the triangle formed is not an isosceles triangle.
4). For a point (4, 2),
Distance between (4, 2) and (-2, -4) = [tex]\sqrt{(4+2)^2+(2+4)^2}=6\sqrt{2}[/tex]
Distance between (4, 2) and (4, -4) = [tex]\sqrt{(4-4)^2+(2+4)^2}=6[/tex]
We find the two sides of this triangle measure 6 units therefore, its an isosceles triangle.
Option (4) will be the third point.
